Fluid Mechanics

JEE Main level, numerical answer, medium

28 of 192

Question 28 of 192

Mercury is filled in a tube of radius 2 cm2 \mathrm{~cm} up to a height of 30 cm30 \mathrm{~cm}. The force exerted by mercury on the bottom of the tube is _____ N.
(Given, atmospheric pressure =105Nm−2=10^5 \mathrm{Nm}^{-2}, density of mercury =1.36×104 kg m−=1.36 \times 10^4 \mathrm{~kg} \mathrm{~m}^{-} 3, g=10 m s−2,π=227)\left.{ }^3, \mathrm{~g}=10 \mathrm{~m} \mathrm{~s}^{-2}, \pi=\frac{22}{7}\right)

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Answer: 177

F=P0 A+ρmghA=105×227×(2×10−2)2+1.36×104×10×(30×10−2)(227×(2×10−2)2)F=51.29+125.71=177 N\begin{aligned} & \mathrm{F}=\mathrm{P}_0 \mathrm{~A}+\rho_{\mathrm{m}} \mathrm{ghA} \\ & =10^5 \times \frac{22}{7} \times\left(2 \times 10^{-2}\right)^2 \\ & +1.36 \times 10^4 \times 10 \times\left(30 \times 10^{-2}\right)\left(\frac{22}{7} \times\left(2 \times 10^{-2}\right)^2\right) \\ & \mathrm{F}=51.29+125.71=177 \mathrm{~N}\end{aligned}

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